codeforces Looksery Cup 2015 H Degenerate Matrix

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于 2017-07-20 16:13:00 发布

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原文链接：http://www.cnblogs.com/wzzkaifa/p/7211798.html

本文介绍了一个关于寻找与给定矩阵最接近的退化矩阵的问题。通过定义矩阵的范数及退化条件，采用二分搜索法求解最小可能的范数值，并提供了完整的C++代码实现。

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The determinant of a matrix 2 × 2 is defined as follows:

$\text{[math]}$

A matrix is called degenerate if its determinant is equal to zero.

The norm ||A|| of a matrix A is defined as a maximum of absolute values of its elements.

You are given a matrix $\text{[math]}$ . Consider any degenerate matrix B such that norm ||A - B|| is minimum possible. Determine||A - B||.

Input

The first line contains two integers a and b (|a|, |b| ≤ 109), the elements of the first row of matrix A.

The second line contains two integers c and d (|c|, |d| ≤ 109) the elements of the second row of matrix A.

Output

Output a single real number, the minimum possible value of ||A - B||. Your answer is considered to be correct if its absolute or relative error does not exceed 10 - 9.

Sample test(s)

input
1 2
3 4

output
0.2000000000

input
1 0
0 1

output
0.5000000000

Note

In the first sample matrix B is $\text{[math]}$

In the second sample matrix B is $\text{[math]}$

这道题能够用二分做，由于要求矩阵最大值的最小值，所以最后A矩阵的每一个元素和B矩阵的每一个元素差值都小于等于一个数时最小，所以从0到10^9枚举差值，然后依据范围二分。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
	int n,m,i,j;
	double l,r,mid,a,b,c,d,a1,a2,b1,b2,c1,c2,d1,d2,t1,t2,s1,s2;
	while(scanf("%lf%lf%lf%lf",&a,&b,&c,&d)!=EOF)
	{
		l=0.0,r=1000000000.0;
		for(i=1;i<=100000;i++){
		   mid=(l+r)/2.0;
		   a1=a+mid;a2=a-mid;
	       b1=b+mid;b2=b-mid;
		   c1=c+mid;c2=c-mid;
		   d1=d+mid;d2=d-mid;
		   t1=min(min(a1*d1,a1*d2),min(a2*d1,a2*d2));
		   t2=min(min(b1*c1,b1*c2),min(b2*c1,b2*c2));
		   s1=max(max(a1*d1,a1*d2),max(a2*d1,a2*d2));
		   s2=max(max(b1*c1,b1*c2),max(b2*c1,b2*c2));
		   if(t1<=s2 && t2<=s1)r=mid;
		   else l=mid;
		}
		double ans=l;
		printf("%.11f\n",ans);
	}
	return 0;
}

转载于:https://www.cnblogs.com/wzzkaifa/p/7211798.html